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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.3.79

In Exercises 59–86, find the derivative of y with respect to the given independent variable.
79. y = θ sin(log₇ θ)

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1
Identify the function y = \(\theta\) \(\sin\)(\(\log\)_{7} \(\theta\)) and recognize that it is a product of two functions: u(\(\theta\)) = \(\theta\) and v(\(\theta\)) = \(\sin\)(\(\log\)_{7} \(\theta\)).
Apply the product rule for derivatives: \(\frac{dy}{d\theta}\) = u'(\(\theta\)) v(\(\theta\)) + u(\(\theta\)) v'(\(\theta\)).
Compute u'(\(\theta\)), the derivative of \(\theta\) with respect to \(\theta\), which is 1.
Find v'(\(\theta\)), the derivative of \(\sin\)(\(\log\)_{7} \(\theta\)). Use the chain rule: the derivative of \(\sin\)(x) is \(\cos\)(x), so multiply by the derivative of \(\log\)_{7} \(\theta\).
Recall that \(\log\)_{7} \(\theta\) = \(\frac{\ln \theta}{\ln 7}\), so its derivative is \(\frac{1}{\theta \ln 7}\). Combine these results to express v'(\(\theta\)) and then substitute back into the product rule formula.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative of Trigonometric Functions

Understanding how to differentiate trigonometric functions like sine is essential. The derivative of sin(u) with respect to x is cos(u) times the derivative of u, applying the chain rule. This allows us to handle compositions involving sine.
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Derivative of Logarithmic Functions with Arbitrary Bases

The derivative of log base a of x, log_a(x), can be expressed using natural logarithms as (1 / (x ln a)). Recognizing this helps differentiate expressions involving logarithms with bases other than e.
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Product Rule for Differentiation

When differentiating a product of two functions, such as θ times sin(log₇ θ), the product rule states that the derivative is the first function times the derivative of the second plus the second function times the derivative of the first. This rule is crucial for correctly finding the derivative.
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